Two Functional Inequalities

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Two Functional Inequalities

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Source: 2025 Nepal Mock TST Day 2 Problem 2

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#1 h Apr 6, 2025, 1:34 PM • 4 Y

Y by PikaPika999, AlexCenteno2007, cubres, khan.academy

Determine all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x) \le x^3$ and $f(x + y) \le f(x) + f(y) + 3xy(x + y)$ for any real numbers $x$ and $y$ .

(Miroslav Marinov, Bulgaria)

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#2 h Apr 6, 2025, 1:45 PM • 2 Y

Y by PikaPika999, cubres

I will show $f(x)=x^3$ for all $x$ , which clearly works.

Taking $x=y=0$ in both, we find $f(0)\le 0$ and $f(0)\ge 0$ , so $f(0)=0$ . Taking $y=-x$ in the second, we find $0=f(0)\le f(x)+f(-x)$ . Likewise, $f(x)\le x^3$ and $f(-x)\le (-x)^3=-x^3$ . So, $f(x)+f(-x)\le 0$ . Combining, $f(x)+f(-x)=0$ for all $x$ . Lastly, using $f(-x)=-f(x)$ , we get $-x^3 \ge f(-x)=-f(x)$ , so $f(x)\ge x^3$ too. Thus, $f(x)=x^3$ for all $x$ .

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#3 h Apr 6, 2025, 1:46 PM • 2 Y

Y by PikaPika999, cubres

Putting $x=y=0$ , we have $f(0)\leq 2f(0)$ , but $f(0)\leq 0$ , so we have to have that $f(0)=0$ .
Putting now $y=-x$ , we have $0\leq f(x)+f(-x) \leq x^3 + (-x)^3 = 0$ , but this means that we have to have $f(x)=x^3$ , because of inequalities.

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#4 h Apr 6, 2025, 4:33 PM • 2 Y

Y by PikaPika999, cubres

Claim: $f(x) \equiv x^3$ is the only function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfies the given conditions.
Let P(x,y) be the assertion for the convenience.
i.e $f(x+y) \le f(x)+f(y)+3xy(x+y)$
P(0,0): $f(0)\ge0$ and $f(0)\le0$ , $\Rightarrow$ $f(0)=0$ ,
Assume $f(x)=g(x)+x^3$ , $\implies$ $g(x+y)\le g(x)+g(y)$
This tells us that g is subadditive.
$g(x)\le 0$ as $f(x)\le x^3$
Let Q(x,y) be the assertion for $g(x+y)\le g(x)+g(y)$
Q(x,-x): $g(0)\le g(x)+g(-x)$ $\rightarrow g(x)+g(-x)\ge 0$ as $f(0)=0$ but $g(x)+g(-x)\le 0$
So, $g(x)+g(-x)=0$ $\Rightarrow g(-x)=-g(x) \ge 0$
The only way both $g(x)\le x$ and $g(-x)\ge 0$ hold is if $g(x)=0$ , $\forall$ $x\in \mathbb{R}$
Therefore $\boxed{f(x)=x^3}$ , $\forall$ $x\in \mathbb{R}$

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#5 h Apr 6, 2025, 7:01 PM • 5 Y

Y by PikaPika999, Assassino9931, megarnie, navier3072, cubres

Let $g(x)=f(x)-x^3+x$ , it reduces to this old problem:
https://artofproblemsolving.com/community/c6h257315p1403063

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#6 h Apr 6, 2025, 7:16 PM • 2 Y

Y by PikaPika999, cubres

Note that $x=y=0 \implies 0 \leq f(0) \leq 0 \implies f(0)=0.$ Now $y=-x \implies 0 \leq f(x)+f(-x) \leq f(x) -x^3 \implies x^3 \leq f(x) \implies f(x)=x^3$ for all $x,$ which clearly works.

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#7 h Apr 6, 2025, 10:18 PM • 1 Y

Y by cubres

jasperE3 wrote:

Let $g(x)=f(x)-x^3+x$ , it reduces to this old problem:
https://artofproblemsolving.com/community/c6h257315p1403063

Nice catch haha

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#8 h Apr 7, 2025, 3:59 AM • 1 Y

Y by cubres

jasperE3 wrote:

Let $g(x)=f(x)-x^3+x$ , it reduces to this old problem:
https://artofproblemsolving.com/community/c6h257315p1403063

Good one!!

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